# The sum of a man's age and his son is 66. What are their ages if the digits are reveresed ? The teacher said there are 3 answers.

Let the man's age be ab and the son's age be ba.

Now we are given that the sum of their ages is 66.

So ab + ba = 66

=> 10a + b + 10b + a = 66

=> 11a + 11b = 66

=> a + b = 6

If a = 6 , b = 0

If a = 5, b = 1

If a = 4, b = 2

For values of a = 3 , 2 , 1 and 0 the son's age is no longer less than the man.

Therefore the age of the man and the son can be

(60, 6), (51 ,15) and (42 , 24)

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Let the age of the father is presented by the 2 digit number xy.

Then, the son's age is the 2 digit number yx.

==> The fathers age = 10x + y

==. The son's age = 10y + x

But we know that the sum of their ages is 66.

==> 10x + y + 10y + x = 66

==> 11x + 11y = 66

We will divide by 11

==> x + y = 6

Then the sum of the digit must be 6.

Let us determine the positive integers whose sums are 6.

==> 0 + 6 = 6

Then the fathers age = 60

and the son's age = 06

Also, : 2 + 4 = 6

==> The father's age = 42

The son's age = 24

Also, 5 + 1 = 6

==> The fathers age = 51

The son's age = 15

Also, 3 + 3 = 6

==> The father's age = 33

The son's age = 33

But this solution is impossible because the father's and son's age can not be the same.

( 51 and 15 ) (24 and 42) and ( 60 and 6)

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